题目
Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string “”.
The testcases will be generated such that the answer is unique.
解题思路
用一个Counter存每个字母所需的个数，使用滑动窗口，当不缺所有的字符的时候，求出窗口的长度，之后如果满足条件就移动左边指针，否则移动右边指针，扫描整个字符串
代码
class Solution:
def minWindow(self, s: str, t: str) -> str:
i = j = I = J = 0
miss = len(t)
need = Counter(t)
for c in s:
if need[c] > 0:
miss -= 1
need[c] -= 1
j += 1
if miss == 0:
while i < j and need[s[i]] < 0:
need[s[i]] += 1
i += 1
if J == 0 or j - i < J - I:
I, J = i, j
return s[I:J]
